By passing a certain amount of cahrge through NaCl solution 9.2 litre of `Cl_(2)` wre liberated at STP. When the same charge is passed through a nitrate solution of a metal M, 7.467 g of the metal was deposited. If the specific heat of the metal is 0.216 cal `g^(-1)`, what is the formula of metal nitrate?

Explanation

By dulong and Petit's law,
Sp, heat`xx`Atomic mass=6.4
`therefore`Atomic mass of the metal`=(6.4)/("Sp. Heat")=(6.4)/(0.216)=29.63`
Suppose the valency of the metal in 'n'. The equivalent wt. of the metal `=(29.63)/(n)`
`22.4L` of `Cl_(2)` have mass=71g
`therefore9.2L` of `Cl_(2)` will have mass`=(71)/(22.4)xx9.2=29.161g`
Applying faraday's second laaw of electrolysis
`("Mass of metal deposited")/("Mass of "Cl_(2)" Liberated")=("Eq. wt. of metal")/("Eq. at. of chlorine")`
`(7.467)/(29.161)=(29.64//n)/(35.5)` or `(29.63)/(n)=(35.5xx7.467)/(29.161)`
or `n=(29.63xx29.161)/(35.5xx7.467)=3.25`.
Taking the nearest whole number, valency of metal=3.
Hence, the formula of metal nitrate will be `M(NO_(3))_(3)`.